| Exam Board | OCR |
|---|---|
| Module | Further Discrete (Further Discrete) |
| Year | 2021 |
| Session | November |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Groups |
| Type | Non-group structures |
| Difficulty | Challenging +1.8 This is a multi-part game theory question requiring understanding of play-safe strategies, stability conditions, dominance reduction, and linear programming formulation. While the individual concepts are A-level Further Maths standard, the question requires synthesizing multiple game theory techniques across four parts, with part (c) requiring algebraic manipulation as a function of x and part (d) requiring knowledge of LP formulation for mixed strategies—more demanding than typical textbook exercises but still within established FM syllabus content. |
| Spec | 7.08a Pay-off matrix: zero-sum games7.08b Dominance: reduce pay-off matrix7.08c Pure strategies: play-safe strategies and stable solutions7.08d Nash equilibrium: identification and interpretation |
| X | Y | Z | ||
| \cline { 3 - 5 } | P | \(x\) | 3 | 2 |
| \cline { 3 - 5 } | Q | 4 | 0 | - 2 |
| \cline { 3 - 5 } | R | - 3 | - 1 | - 3 |
| \cline { 3 - 5 } | ||||
| \cline { 3 - 5 } |
| Answer | Marks | Guidance |
|---|---|---|
| 5 | (a) | (i) |
| Answer | Marks |
|---|---|
| game | M1 |
| Answer | Marks |
|---|---|
| [2] | 1.1 |
| 1.1 | Row minima (P = x, 2; Q = –2; R = –3) |
| Answer | Marks |
|---|---|
| (ii) | Column maxima are max(x, 4), 3, 2 |
| Answer | Marks |
|---|---|
| If -2 x < 2, row maximin = x and col minimax = 2 unstable | M1 |
| Answer | Marks |
|---|---|
| [2] | 1.1 |
| 1.1 | Column maxima, or equivalent (or implied from part (a)(i)) |
| Answer | Marks | Guidance |
|---|---|---|
| 5 | (b) | X Y Z |
| Answer | Marks | Guidance |
|---|---|---|
| Q 4 0 -2 | B1 | |
| [1] | 1.1 | Row R deleted |
| 5 | (c) | If Beth plays X: Alex expects px + 4(1 – p) or 4 + (x – 4)p |
| Answer | Marks |
|---|---|
| 8−𝑥 | M1 |
| Answer | Marks |
|---|---|
| [4] | 3.1a |
| Answer | Marks |
|---|---|
| 1.1 | Calculating expressions for at least two of these |
| Answer | Marks | Guidance |
|---|---|---|
| x | 3 | 2 |
| 5 | (d) | Add 5 (or greater) throughout to make all entries non-negative |
| Answer | Marks |
|---|---|
| and m, p, q, r 0 | B1 |
| Answer | Marks |
|---|---|
| [3] | 3.3 |
| Answer | Marks |
|---|---|
| 3.4 | Add a constant throughout to make all entries |
Question 5:
5 | (a) | (i) | Row minima are min(x, 2), -2, -3
If x 2, row maximin = 2 and col minimax = 2 stable
game | M1
A1
[2] | 1.1
1.1 | Row minima (P = x, 2; Q = –2; R = –3)
Stable when x 2
Allow x > 2
(ii) | Column maxima are max(x, 4), 3, 2
P is a play-safe strategy x -2
If -2 x < 2, row maximin = x and col minimax = 2 unstable | M1
A1
[2] | 1.1
1.1 | Column maxima, or equivalent (or implied from part (a)(i))
Unstable when -2 x < 2
Accept -2 < x < 2 but not just x < 2
5 | (b) | X Y Z
P x 3 2
Q 4 0 -2 | B1
[1] | 1.1 | Row R deleted
5 | (c) | If Beth plays X: Alex expects px + 4(1 – p) or 4 + (x – 4)p
If Beth plays Y: Alex expects 3p
If Beth plays Z: Alex expects 2p – 2(1 – p) or 4p – 2
4p – 2 = px + 4(1 – p)
6
p =
8−𝑥 | M1
A1
M1
A1
[4] | 3.1a
3.1a
3.2a
1.1 | Calculating expressions for at least two of these
All three correct in any form
Graph is optional and carries no marks
Setting their expression for X = 4p – 2 or = 3p
Correct expression for p
x | 3 | 2
5 | (d) | Add 5 (or greater) throughout to make all entries non-negative
0 8 7
9 5 3
2 4 2
Choose row P with probability p, row Q with probability q and row R with
probability r
Maximise M = m – 5
Subject to m 9q + 2r
m 8p + 5q + 4r
m 7p + 3q + 2r
p + q + r 1
and m, p, q, r 0 | B1
B1
B1
[3] | 3.3
3.4
3.4 | Add a constant throughout to make all entries
non-negative
Or using reduced matrix
Need not be stated, may use different notation
e.g. p , p , p
1 2 3
m ap + bq + cr where a, b, c are the values
from a column of their matrix (or the original
matrix if no evidence of change to matrix)
Or m ap + bq using reduced matrix
p + q + r 1 (not p + q + r = 1)
Or p + q 1 using reduced matrix5 Alex and Beth play a zero-sum game. Alex chooses a strategy P, Q or R and Beth chooses a strategy $\mathrm { X } , \mathrm { Y }$ or Z . The table shows the number of points won by Alex for each combination of strategies. The entry for cell $( \mathrm { P } , \mathrm { X } )$ is $x$, where $x$ is an integer.
\begin{table}[h]
\begin{center}
\captionsetup{labelformat=empty}
\caption{Beth}
\begin{tabular}{ l c | c | c | c | }
& & \multicolumn{1}{c}{X} & \multicolumn{1}{c}{Y} & \multicolumn{1}{c}{Z} \\
\cline { 3 - 5 }
& P & $x$ & 3 & 2 \\
\cline { 3 - 5 }
& Q & 4 & 0 & - 2 \\
\cline { 3 - 5 }
& R & - 3 & - 1 & - 3 \\
\cline { 3 - 5 }
& & & & \\
\cline { 3 - 5 }
\end{tabular}
\end{center}
\end{table}
Suppose that P is a play-safe strategy.
\begin{enumerate}[label=(\alph*)]
\item \begin{enumerate}[label=(\roman*)]
\item Determine the values of $x$ for which the game is stable.
\item Determine the values of $x$ for which the game is unstable.
The game can be reduced to a $2 \times 3$ game using dominance.
\end{enumerate}\item Write down the pay-off matrix for the reduced game.
When the game is unstable, Alex plays strategy P with probability $p$.
\item Determine, as a function of $x$, the value of $p$ for the optimal mixed strategy for Alex.
Suppose, instead, that P is not a play-safe strategy and the value of $x$ is - 5 .
\item Show how to set up a linear programming formulation that could be used to find the optimal mixed strategy for Alex.
\end{enumerate}
\hfill \mbox{\textit{OCR Further Discrete 2021 Q5 [12]}}